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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The probaility that a certaun radioactive atom would get disintefrated in a time equal to the mean life fo the radioactive sample isA. `0.37`B. `0.63`C. `0.50`D. `0.67` |
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Answer» Correct Answer - B Required probability `P(t) = (N_(0)(1-e^(-lambdat)))/(N_(0)) = 1-e^(-lambda((1)/(lambda)))= 0.63` |
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| 2. |
If `N_(t) = N_(o)e^(lambdat)` then number of disintefrated atoms between `t_(1)` to `t_(2) (t_(2) gt t_(1))` will ve :-A. `N_(o)[e^(lambdat_(2))-e^(lambdat_(3))]`B. `N_(o)[-e^(lambdat_(2)) - e^(-lambdat_(1))]`C. `N_(o)[e^(-lambdat_(1)) - e^(-lambdat_(2))]`D. None of these |
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Answer» Correct Answer - C `DeltaN? = N_(1) rarr N_(2) = N_(0)[e^(-lambdat_(1)) - e^(-lambdat_(2))]` |
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| 3. |
Identify the correct variation of potential energy `U` as a function of displacement `x` from mean position (or `x^(2)`) of a harmonic oscillator (`U` at mean position `= 0`)A. B. C. D. None of these |
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Answer» Correct Answer - C `U = (1)/(2)kx^(2)` i.e. `U` versus `x^(2)` graph is a straight line passing through origin. |
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| 4. |
The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is A. `1 xx 10^(2)Nm^(-1)`B. `1.5 xx 10^(2) Nm^(-1)`C. `2 xx 10^(2)Nm^(-1)`D. `3 xx 10^(2)Nm^(-1)` |
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Answer» Correct Answer - B (b) `(1)/(2) kA^(2)=(0.04-0.01)J=0.03 J` `therefore k=(0.06)/(A^(2))=(0.06)/((0.02)^(2))=150 Nm^(-1)` |
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| 5. |
The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is A. ` 1 xx10^(2) N/m `B. 150N/mC. `667 xx 10^(2) N//m`D. `3 xx 10^(2)` N/m |
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Answer» Correct Answer - B |
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| 6. |
A body performs S.H.M. Its kinetic energy K varies with time t as indicated by graphA. B. C. D. |
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Answer» Correct Answer - A |
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| 7. |
A ball of mass `m kg` hangs from a spring of spring constant `k`. The ball oscillates with a period of `T` seconds if the ball is removed, the spring is shortened(`w.r.t.` length in mean position) byA. `(gT^(2))/((2pi)^(2))` metreB. `(3T^(2)g)/((2pi)^(2))` metreC. `(Tm)/(k,)` metreD. `(Tk)/(m)` metre |
| Answer» Correct Answer - A | |
| 8. |
A smooth inclined plane having angle of inclination `30^(@)` with horizontal has a mass `2.5 kg` held by a spring which is fixed at the upper end as hwon in figure. If the mass is taken `2.5 cm` up along the surface of the inclined plane, the tension in the soring reduces to zero. If the mass is then released, the angular frequency of oscillation in radian per second is A. `0.707`B. `7.07`C. `1.414`D. `14.14` |
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Answer» Correct Answer - D `kx = mg sin 30^(@)` `omega = sqrt((k)/(m)) = sqrt((gsin30^(@))/(x)) = sqrt((5 xx 100)/(2.5)) = 14.14` Ans. |
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| 9. |
A smooth inclined plane having angle of inclination `30^(@)` with horizontal has a mass `2.5 kg` held by a spring which is fixed at the upper end as hwon in figure. If the mass is taken `2.5 cm` up along the surface of the inclined plane, the tension in the soring reduces to zero. If the mass is then released, the angular frequency of oscillation in radian per second is A. `0.707`B. `7.07`C. `1.414`D. `14.14` |
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Answer» Correct Answer - D `kx = mg sin 30^(@)` `omega = sqrt((k)/(m)) = sqrt((gsin30^(@))/(x)) = sqrt((5 xx 100)/(2.5)) = 14.14` Ans. |
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| 10. |
A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided in two equal parts and one part isused to continue the simple harmonic motion, the time period willA. remain`T`B. becomes `2T`C. become `T//2`D. become `T//sqrt(2)` |
| Answer» Correct Answer - D | |
| 11. |
A paricle of mass `200 g` executes a simple harmonic motion. The restorting force is provided by a spring of spring constant `80 N//m`. Find the time period. |
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Answer» The time period is `T=2pi sqrt(m/k)` `=2pi(sqrt(200x10^-3kg)/(80Nm^-1))` `2pix0.05s=0.31s` |
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| 12. |
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring isA. 4 Mg/ KB. 2 Mg/ kC. Mg/ KD. Mg/ 2K |
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Answer» Correct Answer - B |
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| 13. |
Two blocks of masses 3 kg and 6kg rest on horizontal smooth surface. The 3 kg block is attached to a spring with a long constant `k=900Nm^(-1)` which is compressed 2m from beyond 1m from mean position. 3kg mass strikes the 6 kg mass and the two stick together. A. Veocity of the combined masses immediately after the collision is `10ms^(-1)`.B. velocity of the combined masses immediately after the collision is `5ms^(-1)`C. amplitude of the resulting oscillation is `sqrt(2)`m.D. amplitude of the resulting oscillation is 1m. |
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Answer» Correct Answer - A::C |
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| 14. |
A cord is attached between a 0.50 kg block and a string with force constant `k=20 N//m`.The other end of the spring is attrached to the wall and the cord is placed over a pulley `(I=0.60 MR^2)` of mass 5.0 kg and radius 0.50 m.(See the accompanying figure.) Assuming no slipping occurs what is the frequency of the oscillations when the body is set into motion ? |
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Answer» Correct Answer - C In equilibrium `kx_(0) = mg` `:. x_(0) = (mg)/(k) = (0.5 xx 10)/(20)` `= 0.25m` When displaced downwards by `x` from the mean position, total mechanical energy, `E = - mgx + (1)/(2)mv^(2) + (1)/(2)Iomega^(2) + (1)/(2)(x + x_(0))^(2)` Substituting `I = 0.6 MR^(2)` and `omega = (v)/(R)` We have, `E = - mgx + (1)/(2)mv^(2) + (1)/(2) (0.6MR^(2)) ((v)/(R))^(2) + (1)/(2)k (x + x_(0))^(2)` Since E = constant `:. (dE)/(dt) = 0` or `0 = - mg ((dx)/(dt)) + (1)/(2)m (2v.(dv)/(dt)) + (0.6M) + (v)(dv)/(dt) + (1)/(2)k[2(x + x_(0))] (dx)/(dt)` putting `(dx)/(dt) = v, kx_(0) = mg` and `(dv)/(dt) = a` We have `0 = (ma + 0.6 Ma) + kx` or `a = - ((k)/(m + 0.6M))x` Since, `a prop - x` motion is SHM. `f = (1)/(2pi) sqrt|(a)/(x)|` `= (1)/(2pi)sqrt((k)/(m + 0.6M))` ` = (1)/(2pi) sqrt((20)/(0.5 xx 0.6 xx 5))` ` = 0.38 Hz` . |
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| 15. |
A point particle of mass `0.1kg` is executing `SHM` with amplitude of `0.1m`. When the particle passes through the mean position. Its `K.E.` is `8xx10^(-3)J`. Obtain the equation of motion of this particle if the initial phase of oscillation is `45^(@)`. |
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Answer» Correct Answer - `Y=0.1SIN(4T+PI//4)` |
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| 16. |
A rubber cord of force constant k = 100 N/m and l=1m is attached to a particle of mass m = 1kg at one end and fixed to a vertical wall at the other. The body is displaced by `x_(0)` = 25cm so as to stretch the cord and then released. Calculate the time particle takes to reach the wall. |
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Answer» Correct Answer - `sqrt((m)/(k))((pi)/(2)+(l)/(x_(0)))=0.56s` |
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| 17. |
The graph in the figure shows how the displacement of a particle describing S.H.M. varies with time. Which one of the following statements is not true ? A. the force is zero at time `(3T)/4`B. the velocity is maximum at time T/2C. the acceleration is maximum at time TD. the P.E. = total energy at the time T/2 |
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Answer» Correct Answer - B The velocity is maximum at time T/2 |
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| 18. |
A string of length l=1m and tensioned by a weight M = 4kg has two masses, each of mass = 10g, attached to it at distance `l/4` and `3/4` from one of the ends. Find the frequency of oscillations of the masses in the two cases. a) When the masses are displaced equally in the same direction b) when the masses are displaced equally in opposite directions. |
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Answer» Correct Answer - `(a)(1)/(2pi)sqrt((4Mg)/(ml)),(b)(1)/(2pi)sqrt((8Mg)/(ml))` |
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| 19. |
A block of mass `m_(1)=1kg` is attached to a spring of force constant `k=24N//cm` at one end and attached to a string tensioned by mass `m_(2)=5kg` . Dedcue the frequency of oscillaitons of the system. If `m_(2)` is initially supported in hand and then suddenly released, find `(a)` instantaneous tension just after `m_(2)` is released . ltbr. `(b)` the maximum displacement of `m_(1)`. `(c)` the maximum and minimum tensions in the string during oscillations. |
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Answer» Correct Answer - `v=(1)/(2pi)sqrt((K)/(m_(1)+m_(2)))=(10)/(2x)Hz,m_(2)delta,(2m_(2)g)/(k),T_("max")=(55g)/(6),T_("min")=(5g)/(6)` |
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| 20. |
A particle of mass `4 kg` moves between two points `A` and `B` on a smooth horizontal surface under the aciton of two forces such that when it is at a point `P`, the forces are `2vec(P)AN` and `2vec(P)B N`. If the particle is released from rest at `A,` find the time it takes to travel a quarter of the way `A` to `B`. |
| Answer» Correct Answer - `(pi)/(3)s` | |
| 21. |
In the figure shown, `m_(a)=m_(b)=1kg` block-`A` is neutral while block-`B` carries charge `-1c`. Sizes of `A` and `B` are negligible. Block-`B` is released from rest at a distance `1.8m` from `A`. Initially spring is neither compressed nor stretched. If collision between `A` and `B` is perfectly inelastic, then velocity of the combined mass just after collision isA. `6m//s`B. `3m//s`C. `9m//s`D. `12m//s` |
| Answer» Correct Answer - B | |
| 22. |
A small bob attached to a light inextensible thread of length `l` has a periodic time `T` when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end `O` of a vertical rigid rod of length `(3l)/(4)` (as in figure). If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be- A. `(3T)/(4)`B. `(T)/(2)`C. `T`D. `2T` |
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Answer» Correct Answer - A |
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| 23. |
A small bob attached to a light inextensible thread of length `l` has a periodic time `T` when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end `O` of a vertical rigid rod of length `3l//4`. If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be A. `3T//4`B. `4T//5`C. `2T//3`D. `5T//6` |
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Answer» Correct Answer - A Half of the oscillation is completed with length `l` and rest half with `l//4`. `:.` Time period `= (T_(1))/(2) + (T_(2))/(2)` `= (1)/(2)[2pi sqrt((l)/(g)) + 2pi sqrt((l//4)/(g))]` `= (3)/(4)[2pi sqrt((l)/(g))] = (3)/(4)T` |
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| 24. |
A simple pendulum of length l has a brass bob attached at its lower end. Its period is T . If a steel bob of same size, having density x times that of brass, replaces the brass bob and its length is changed so that period becomes 2 T, then new length isA. 2lB. 4 lC. `4 lx`D. `(4l )/x` |
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Answer» Correct Answer - B |
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| 25. |
Time period of a simple pendulum of length L is `T_(1)` and time period of a uniform rod of the same length L pivoted about an end and oscillating in a vertical plane is `T_(2)`. Amplitude of osciallations in both the cases is small. Then `T_(1)/T_(2)` isA. `sqrt(4/3)`B. 1C. `sqrt(3/2)`D. `sqrt(1/3)` |
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Answer» Correct Answer - C |
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| 26. |
If the length of a pendulum is made 9 times and mass of the bob is made 4 times then the value of time period becomesA. 3TB. 3/2 TC. 4 TD. 2T |
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Answer» Correct Answer - A |
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| 27. |
A simple pendulum is oscillating in a lift. If the lift is going down with constant velocity, the time period of the simple pendulum is `T_(1)`. If the lift is going down with some retardation its time period is `T_(2)`, thenA. `T_(1)gtT_(2)`B. `T_(1)ltT_(2)`C. `T_(1)=T_(2)`D. depends upon the mass of the pendulum bob |
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Answer» Correct Answer - A |
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| 28. |
A simple pendulum is oscillating in a lift. If the lift starts moving upwards with a uniform acceleration, the period willA. remain unaffectedB. be shorterC. be longerD. may be shorter or longer, depending on the magnitude of acceleration |
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Answer» Correct Answer - B `T=2pisqrt((L)/(g+a))` |
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| 29. |
A particle executes simple harmonic motion with an amplitude of `4cm` At the mean position the velocity of the particle is `10`cm/s. distance of the particle from the mean position when its speed `5` cm/s isA. `sqrt(3)` cmB. `sqrt(5)` cmC. `2sqrt(3)` cmD. `2sqrt(5)` cm |
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Answer» Correct Answer - C `v_("max")=aomegaimpliesomega=(v_("max"))/(a)=(10)/(4)` Now, `v=omegasqrt(a^(2)-y^(2))` `implies v^(2)=omega^(2)(a^(2)-y^(2))` `implies y^(2)=a^(2)-(v^(2))/(omega^(2))` `implies y=sqrt(a^(2)-(v^(2))/(omega^(2)))=sqrt(4^(2)-(5^(2))/((10//4)^(2)))=2sqrt(3)` cm |
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| 30. |
In gamma ray emission from a nucleusA. both the neutron number and the proton number changeB. there is no change in the proton number and the neutron numberC. only the neutron number of changesD. only the proton number changes |
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Answer» Correct Answer - B Gamma ray is an electronagentic radiation, dust to the emission of gamma ray, neither the mass number not the atomic number changes. Though the daughter nucleus is same as parent nucleus but still there is a difference in the two, i.e., the daughter nucleus so obtained is present in one of the exclted states and not in the ground state. |
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| 31. |
`.^(12)N` beta decays to an ecited state of `.^(12)C`, which subsequenctly decays to the ground state with the emission of a `4.43 -MeV` gamma ray. What is the maxium kinetic energy of the emitted beta particle? |
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Answer» To determine the Q value for this decay, we first need to find the mass of the product nucleus `.^(12)C` in its excited state. In the ground state, `.^(12)C` has a mass of `12.000000u`, so its mass in the excited state is `12.000000u + (4.43 MeV)/(931.5 MeV//u) = 12.004756 u` In this decay, a proton is converted to a neutron, so it must be an example of position decay. The Q value is, accoding to equation `Q = (12.018613 u - 12.004756 u - 2 xx 0.000549 u) (931.5 MeV//u)` `= 11.89 MeV` (We could have just as easily found the Q value by first finding the Q value for decay to the ground state, `16.32 MeV`, and the subtracting the excitation energy of `4.43 MeV` since the decay to the excited state has that much less availbale energy.) Neglecting the small correction for the recoil kinetic energy of the `.^(12)C` nucleus, the maximum electron kinetic energy is `11.89 MeV`. |
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| 32. |
Which of the following process represents a `gamma- decay`?A. `.^(A)X_(Z) + gamma rarr .^(A)X_(Z-1) + a + b`B. `.^(A)X_(Z) + .^(1)n_(0) rarr .^(A-3)X_(Z-2) + o`C. `.^(A)X_(Z) rarr .^(A)X_(Z) + f`D. `.^(A)X_(2)+e_(-1)rarr .^(A)X_(A-1)+g` |
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Answer» Correct Answer - C During `gamma-` decays atomic number (Z) and mass number (A) does not change. So, the correct option is (c) because in all other options either Z, A or both is/are changing. |
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| 33. |
A cylinderical block of wood of mass m and area cross-section A is floating in water (density = `rho`) when its axis vertical. When dressed a little and the released the block starts oscillating. The period oscillations isA. `(2pisqrt(m/(rhobidwedgeg))`B. `2pisqrt((mg)/(rhoA))`C. `2pisqrt((rhoAg)/(m))`D. `2pisqrt(rhoA)/(mg))` |
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Answer» Correct Answer - A |
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| 34. |
The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:A. `540 nm`B. `400 nm`C. `310 nm`D. `220 nm` |
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Answer» Correct Answer - C `lambda` (in Å)`=(12375)/(W(eV))=12375/4.0 Å~~3093 Å` `implies lambda~~309.3 nmimplies lambda=310 nm` |
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| 35. |
The half life of `I""^(131)` is 8 day. Given a sample of `I""^(131)` at t=0, we can assert thata)No nucleus will decay at t=4dayb)No nucleus will decay before t=8dayc)All nucleus will decay before t=16dayd)A given nucleus may decay beforeA. no nucleus will decay before `t = `4 daysB. no nucleus will decay before `t = 8` daysC. all nuclei will decay before `t = 16` daysD. a given nucleus may decay at any time after `t = 0` |
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Answer» Correct Answer - D Number of nuclei decreases experientially `N=N_(0)e^(-lambdat)` and Rate of decay `(-(dN)/(dt))=lambdaN` Therefore, decay process lasts upto `t=oo`. Therefore a given nucleus may decay at any time after `t=0`. |
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| 36. |
A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength ` lambda = 5.76 xx 10^(-15)m ` if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is `4.002amu` , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 `MeV//e^(2)`) |
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Answer» Correct Answer - `6.25 MeV, 227.62 "amu"` (i) Given mass of `alpha`-particle, `m = 4.002` amu and mass of daughter nucleus `M = 223.610` amu, de-Broglie wavelength of `alpha`-particle, `lambda = 5.76 xx 10^(-15) m` So, momentum of `alpha`-particle would be `p = (h)/(lambda) = (6.63 xx 10^(-34))/(5.76 xx 10^(15)) kg- m//s` `rArr = 1.151 xx 10^(-19) kg - m//s` From law of conservation of linear momentum, this should also be equal to the linear momentum the daughter nucleus (in opposite direction). Let `K_(1)` and `K_(2)` be the kinetic energies of `alpha`-particle and daughter nucleus. Then total kinetic energy in the final state is `K = K_(1) + K_(2) = (p^(2))/(2m) + (p^(2))/(2M)` `= (p^(2))/(2) ((1)/(m) + (1)/(M)) = (o^(2))/(2) ((M + m)/(Mm))` 1 amu `= 1.67 xx 10^(-27) kg` Substituting the values, we get `K = ((1.151 xx 10^(-19))^(2))/(2) xx ((M + m)/(M xx m))` `= (p^(2))/(2) xx ((4.002 + 223.61) (1.67 xx 10^(-27)))/((4.002 xx 1.67 xx 10^(-27)) (223.61 xx 1.67 xx 10^(-27)))` `K = 10^(-12) J = (10^(-12))/(1.6 xx 10^(-13)) MeV = 6.25 MeV` `rArr K = 6.25 MeV` (ii) Mass defect, `Deltam = (6.25)/(931.470)` amu `= 0.0067` amu Therefore, mass of parent nucleus `=` mass of `alpha`-particle `+` mass of daughter nucleus `+` mass defect `(Delta m)` `= (4.002 + 223.610 + 0.0067)` amu `= 227.62` amu |
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| 37. |
When a nucleus with atomic number `Z` and mass number `A` undergoes a radioactive decay process, (i) Both `Z` and `A` will decrease, if the process is `alpha` decay (ii) `Z` will decrease but `A` will not change, if the process is `beta^(+)-`decay (iii) `Z` will increase but `A` will not change, if the process is `beta-`decay (iv) `Z` and `a` will remain uncharged, if the prices is `gamma` decayA. both Z and A will decrease, If the process is `alpha` decayB. Z will decrease but A will not change, if the process is `beta^(+)` decayC. Z will decrease but A will not change, if the process is `beta^(-)` decayD. Z and A will remain unchanged, if the process is `beta` decay |
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Answer» Correct Answer - A::B::D `alpha` decays : `._(2)He^(4)`, so both Z & A decreases. `beta^(+)` decays : `._(+1)e^(0)`, so A will not change but Z will change (decreases) `beta^(-)` decat : `._(-1)e^(0)`, so A will not change but Z will change (increases) `gamma` decays : no change in A & Z. |
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| 38. |
The isotope `_(5)^(12) B` having a mass `12.014 u`undergoes beta - decay to `_(6)^(12) C _(6)^(12) C `has an excited state of the nucleus `( _(6)^(12) C ^(**) at 4.041 MeV` above its ground state if `_(5)^(12)E` decay to `_(6)^(12) C ^(**) ` , the maximum kinetic energy of the `beta` - particle in unit of MeV is `(1 u = 931.5MeV//c^(2)` where c is the speed of light in vaccuum) . |
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Answer» Correct Answer - 9 `._(2)^(12)B rarr ._(6)^(12) C+ ._(-1)^(0)e+bar(v)` Mass defect `=(12.014-12) u` `:.` Released energy `=13.041 MeV` Energy used for excitation of `._(6)^(12)C=4.041 MeV` `:.` Energy converted to KE of electron `=13.041-4.041=9` MeV |
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| 39. |
Statement-1 : Pendulum clocks go slow in summer and fast in winter. Statement-2 : The length of the pendulum used in clock increases in summer.A. Statement-1 is True, Statement-2 is True , Statement-2 is a corrrect explanation for Statement-1B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a corrrect explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 40. |
A vertical mass-spring system executed simple harmonic oscillation with a period `2s` quantity of this system which exhibits simple harmonic motion with a period of `1sec` areA. VelocityB. Potential energyC. Phase difference between acceleration and displacemenD. Difference between kinetic energy and potential energy |
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Answer» Correct Answer - A |
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| 41. |
A simple pendulum , a physical pendulum, a torsional pendulum and a spring-mass systeam, each of same frequency are taken to the Moon. If frequencies are measured on the moon, which systeam of systeams will have it unchanged?A. spring-mass system and torisonal pendulum.B. only spring-mass system.C. spring-mass system and physical pendulum.D. None of these |
| Answer» Correct Answer - A | |
| 42. |
A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance `(3l)/4`. The pendulum is released from rest. Throughout the motion, the moving string makes small angle with vertical. Time after which the pendulum returns back to its initial position is A. `T`B. `(3T)/(4)`C. `(3T)/5`D. `(4T)/5` |
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Answer» Correct Answer - B |
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| 43. |
Pendulum `A` is a physical pendulum made from a thin, rigid and uniform rod whose length is `d`. One end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. Pendulum `B` is a simple pendulum whose length is also `d`. Obtain the ratio `(T_(A))/(T_(B))` of their periods for small angle oscillations.A. `sqrt((3)/(2))`B. `sqrt((2)/(3))`C. `(2)/(3)`D. `3/2` |
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Answer» Correct Answer - B `T_(A) = 2pisqrt((I)/(mgl)) = 2pisqrt((mgl^(2)//3)/(mgl//2)) = 2pisqrt((2l)/(3g))` & `T_(B) = 2pisqrt((l)/(g)) rArr (T_(A))/(T_(B)) = sqrt((2)/(3)` |
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| 44. |
A rigid body is to be suspended like a physical pendulum so as to have a time period of `T = 0.2pi` second for small amplitude oscillations. The minimum distance of the point of suspension from the centre of mass of the body is `l_(1) = 0.4 m` to get this time period. Find the maximum distance `(l_(2))` of a point of suspension from the centre of mass of the body so as to get the same time period. `[g = 10 m//s^(2)]` |
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Answer» Correct Answer - 0.6 m |
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| 45. |
Pendulum `A` is a physical pendulum made from a thin, rigid and uniform rod whose length is `d`. One end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. Pendulum `B` is a simple pendulum whose length is also `d`. Obtain the ratio `(T_(A))/(T_(B))` of their periods for small angle oscillations. |
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Answer» Correct Answer - A `T_(A) = 2pi sqrt ((I)/(mgl))` `= 2pi sqrt (((md^(2)//3))/(mg(d//2))) = ((sqrt2)/(3))2pi sqrt((d)/(g))` and `T_(B) = 2pi sqrt((d)/(g))` `:. (T_(A))/(T_(B)) = sqrt((2)/(3)) = 0.816` |
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| 46. |
A particle is executing SHM given by `x = A sin (pit + phi)`. The initial displacement of particle is `1 cm` and its initial velocity is `pi cm//sec`. Find the amplitude of motion and initial phase of the particle. |
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Answer» Correct Answer - A::B::C::D `x = A sin (pit + phi)` At `t = 0 , x= 1 cm rArr A sinphi = 1"…."(i)` Velocity `v = (dx)/(dt) = piA cos (pit + phi)` At `t = 0 pi = piA cos phi rarr Acosphi = 1"……."(ii)` from `(i)` & `(ii) A^(2) (sin^(2)phi + cos^(2) phi) = 1 + 1` `rArr A = sqrt(2) cm` and `tanphi = 1 rArr phi = (pi)/(4) rad` |
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| 47. |
In `SHM` there is always a constant ratio between displacement of the body and itsA. velocityB. accelerationC. massD. time period |
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Answer» Correct Answer - B `a=-omega^(2)x` `therefore (x)/(2)=-(1)/(omega^(2))`=constant |
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| 48. |
In the problem if k = 10 N/m, m=2kg, R=1m and A=2m. Find linear speed of the disc at mean position. A. `sqrt(40/3)`m/sB. `sqrt(20)`m/sC. `sqrt(10/3)`m/sD. `sqrt(50/3)`m/s |
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Answer» Correct Answer - A |
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| 49. |
In the problem if k = 10 N/m, m=2kg, R=1m and A=2m. Find linear speed of the disc at mean position. A. `sqrt((40)/(3)) m//s`B. `sqrt(20) m//s`C. `sqrt((10)/(3)) m//s`D. `sqrt((50)/(3)) m//s` |
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Answer» Correct Answer - A `1/(2)kAY^(2) = 1/2 mv^(2) (1+(k^(2))/(R^(2))) = 1/2mv^(2)(1+1/2) = 3/4 mv^(2)` `v = sqrt((2kA^(2))/(3m)) = sqrt((2(10)(2)^(2))/(3(2))) = sqrt((40)/(3))ms^(-1)` |
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| 50. |
An object of mass m is moving in unifrom circular motion in xy-plane. The circle has radius R and object is moving clockwise around the circle with speed v. The motion is projected onto the x-axis where it appears as simple harmonic motion accoding to `x(t) = Rcos (omegat + phi)`. The motion starts from pointof coordinates `(0,R)`. In this projection `phi` is-A. `pi//2`B. `pi`C. `3pi//2`D. `0` |
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Answer» Correct Answer - C At `t = 0, x = 0 , v = v_(0) = omegaR` so `cos(omegat + phi) = 0` & `-sin(omegat+phi) = +1 rArr phi = (3pi)/(2)` |
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